Normal Matrices — Mathematics Knowledge System

Overview

A matrix A is normal if it commutes with its adjoint: AA* = A*A. Normal matrices are exactly those that are unitarily diagonalizable — they can be written A = U D U* where U is unitary and D is diagonal. This class includes Hermitian, skew-Hermitian, and unitary matrices, and the spectral theorem for normal matrices is the cornerstone of their theory.

Intuition

Normal matrices are the 'nice' matrices of the complex world — they have a complete orthonormal basis of eigenvectors. The condition AA* = A*A says the matrix and its adjoint do not interfere with each other. Hermitian matrices (A = A*) and unitary matrices (A*A = I) are special cases. If a matrix is not normal, its eigenvectors may not span the space, requiring Jordan form instead of diagonalization.

Formal Definition

Definition

A matrix A ∈ M_n(C) is normal if it commutes with its conjugate transpose:

A \text{ normal} \iff AA^* = A^*A

Normality condition

A = U D U^*, \quad U \text{ unitary}, \quad D = \operatorname{diag}(\lambda_1,\ldots,\lambda_n)

A is normal iff it is unitarily diagonalizable.

Spectral theorem for normal matrices

Notation

Notation	Meaning
$A^*$	Conjugate transpose (adjoint) of A
$AA^* = A^*A$	Commutativity with adjoint — the normality condition

Properties

Hermitian matrices are normal

A = A^* \implies AA^* = A^*A. \text{ (Eigenvalues are real.)}

Unitary matrices are normal

A^*A = I \implies AA^* = I = A^*A. \text{ (Eigenvalues satisfy } |\lambda|=1.)

Skew-Hermitian matrices are normal

A = -A^* \implies AA^* = (-A^*)(-A^*)^* = A^*A. \text{ (Eigenvalues are purely imaginary.)}

Orthogonal eigenvectors

\text{Eigenvectors of a normal matrix for distinct eigenvalues are orthogonal.}

Worked Examples

1
Compute A*: take conjugate transpose.
$A^* = \begin{pmatrix}2 & -1-i \\ 1-i & 3\end{pmatrix}$
2
Compute AA*.
$AA^* = \begin{pmatrix}2 & 1+i \\ -1+i & 3\end{pmatrix}\begin{pmatrix}2 & -1-i \\ 1-i & 3\end{pmatrix}$
3
(AA*)_{11} = 4 + (1+i)(1-i) = 4 + 2 = 6; (A*A)_{11} = 4 + (-1-i)(-1+i) = 4 + 2 = 6.
$(AA^*)_{12} = -2(1+i)+3(1+i) = 1+i; \quad (A^*A)_{12} = -2(1+i)+3(1+i) = 1+i$
4
Checking all entries shows AA* = A*A.

✓ Answer

AA* = A*A, so A is normal.

Practice Problems

Mediumfree response

Is the real matrix A = [[1,2],[-2,1]] normal? Show work.

Mediumfree response

State the spectral theorem for normal matrices and describe what it means geometrically.

Common Mistakes

Common Mistake

Every diagonalizable matrix is normal.

A diagonalizable matrix need not be normal — it may require a non-orthogonal change of basis. Normal matrices are UNITARILY diagonalizable, which is a stronger condition.

Quiz

Summary

A matrix A is normal if AA* = A*A (it commutes with its adjoint).
Normal matrices are exactly those unitarily diagonalizable: A = UDU* with U unitary.
Hermitian, skew-Hermitian, and unitary matrices are all special cases of normal matrices.
Eigenvectors of a normal matrix for distinct eigenvalues are orthogonal.
Non-normal matrices may lack an orthonormal eigenbasis and require Jordan form for full analysis.

References

WebsiteWikipedia — Normal matrix